Add two numbers

Question

Input: Two integers. Preferably decimal integers, but other forms of numbers can be used. These can be given to the code in standard input, as arguments to the program or function, or as a list.

Output: Their sum. Use the same format for output integers as input integers. For example, the input 5 16 would lead to the output 21.

Restrictions: No standard loopholes please. This is code-golf, answer in lowest amount of bytes wins.

Notes: This should be fairly trivial, however I'm interested to see how it can be implemented. The answer can be a complete program or a function, but please identify which one it is.

Test cases:

1 2 -> 3
14 15 -> 29
7 9 -> 16
-1 8 -> 7
8 -9 -> -1
-8 -9 -> -17

Or as CSV:

a,b,c
1,2,3
14,15,29
7,9,16
-1,8,7
8,-9,-1
-8,-9,-17

Leaderboard

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Answer 1

Bash, 18 bytes

f(){expr $1 + $2;}

Answer 2

Prelude, 4 bytes

??+!

Requires my modified Prelude interpreter which uses decimal I/O.

Like several other answers, this is just read, read, add, write.

Answer 3

Befunge - 5 Bytes

&&+.@

& - Request int from user and push to stack

+ - Pop top two elements from stack and add and push result

. - Pop value and output as int

@ - End program

Try it here (Although you will have to copy/paste into the text area)

Edit: oops, didn't notice the earlier (identical) befunge answer, I'll leave this here unless I'm told to delete it, not sure of the opinion on that.

Answer 4

GO, 29 bytes

func (a,b int)int{return a+b}

Not that much to say

Answer 5

Yup, 6 bytes

0*-*-#

Try it online!

Explanation

0*-*-#
0        push 0 to the stack   [0]
 *       place input           [0 a]
  -      subtract              [-a]
   *     place input           [-a b]
    -    subtract              [-a-b]
                             = [b+a]
     #   output

As a function on -cheat mode, 9 bytes:

0~--0~-=+

Try it online!

Answer 6

SML, 3 bytes

op+

This is the prefix version of the infix +. If we input it like this in an interpreter (for example Moscow ML), it's type is displayed

- op+;
> val it = fn : int * int -> int

which tells us how to use it: Given a tuple of integers, an integer will be returned.

- op+(17,25);
> val it = 42 : int 

Answer 7

GAP, 2 Bytes

\+

The backslash is GAP's way to turn an infix operator (and some more, there is also \[\] for indexing) to a function.

Here is an example use:

gap> \+(4,3);
7

Answer 8

jq, 3 characters

add

Sample run:

bash-4.3$ jq 'add' <<< '[5,16]'
21

On-line test

Answer 9

Elixir, 7 bytes

& &1+&2

Anonymous function defined using the capture operator. Another version is &(&1+&2), however this approach saves 1 byte. The verbose version is fn a,b->a+b end - 15 bytes.

Full program with test case (yes, the . in the function call is mandatory!):

s=& &1+&2
IO.puts s.(1,6) #7

Try it online on ElixirPlayground !

Answer 10

C, 49 bytes

main(a,b){scanf("%d %d",&a,&b);printf("%d",a+b);}

Can't really do much to golf it down.

For fun:

The following works only if 0 <= sum < 256 (it's 59 bytes long):

a;main(c,v)char**v;{return a++&2?c-3:main(c+atoi(*++v),v);}

Use gcc to compile, and ./a.out [your 2 nums]; echo $? to run it.

Here's the ungolfed version of that program:

/* Global int auto-initialized to 0 */
a;

/* The main method with two int params */
/* 'c' is argc, and 'v' is argv */
main(c, v)

/* Yes, this is valid. It defines 'v' as a char** */
char** v;

{
    /* Checks to see if a == 2, and increments a.
     * I only want recursion to happen twice. */
    if (a++ & 2)
    {
        /* Since "argc" is 3, we need to subtract it from the final answer */
        return c - 3;
    }
    else
    {
        /* Get the next int value of "argv" and add it to 'c'.
         * Then call main() again with the updated value */
        return main(c + atoi(*++v), v);
    }
}

So, main() returns the value of the sum of the two numbers (essentially, I'm forcing it to return a custom exit code). The program won't output the exit code, so calling echo $?, in the same command as running the program, outputs the return value of that program.

The range of exit codes only exist between 0 and 255, so if you run the program trying to sum 255 and 1, it will wrap around and output 0.

Answer 11

Ruby, 17 bytes

** Edit: Big thank you to @steenbergh and @TùxCräftîñg for their help **

def c(a,b)a+b end

Answer 12

Befunge, 7 bytes

#v&<
 +
 .
 @

Befunge is an esoteric language that can execute in any cardinal direction on a 2 dimensional plane. Executing left to right from the top left (default) we get the following set of operations ((x,y) = coordinates of operation with (0,0) in the top left)

(0,0): skip next cell in execution path (1,0)
(2,0): ask for user input and push it to the stack
(3,0): reverse direction of execution
(2,0): ask for user input and push it to the stack
(1,0): begin executing downward
(1,1): pop first two values on stack, add them and push result
(1,2): pop stack and output value
(1,3): end

&&+.@ would also work, and in only 5 bytes, but is not nearly as cool

Answer 13

Swift, 1 byte

+

since operator + is defined as function(public func +(lhs: Int, rhs: Int) -> Int)

Answer 14

C, 66 bytes

Typical C omissions with regard to declarations, header files, and return statements. This actually works for any number of decimal integers (up to the limitations of one's shell), including none. The program can be invoked with ./a.out 1 2 [...], as an example.

main(c,v,x)char**v;{x=0;while(c-1)x+=atoi(v[--c]);printf("%d",x);}

Or, more legibly:

main(c,v,x) char**v; {
    x=0;
    while(c-1) x+=atoi(v[--c]);
    printf("%d",x);
}

Answer 15

QBIC, 6 bytes

::?a+b

It takes two cmd line params as numbers (with two :'s), names them a and b, and adds them while printing (?).

Answer 16

Euphoria, 64 bytes

Code:

include get.e
print(1,prompt_number("",{})+prompt_number("",{}))

Answer 17

Whirl, 32 bytes

01100100001100011110001000111100

Try it online!

Explanation

Whirl has two "rings" of operations, one for control and I/O operations and the other for math operations. The 1 command rotates the current operation ring while the 0 command switches direction of the rotation. Two 0s in a row run the current operation and switch to the other ring.

01100    Read an integer from stdin and store it at the memory pointer
100      Load that value into the math ring data storage
00       Read an integer from stdin and store it at the memory pointer
1100     Add the the values under the memory pointer to the math ring data storage
0111100  Set the control ring data storage to 1
0100     Save the math ring data storage to the memory at the memory pointer
0111100  Print the result from the memory at the memory pointer

Answer 18

V, 2 bytes

À

Try it online!

You can't see it, but after the À there is a control-a character (ASCII 0x01).

This answer is nice because it shows one of the nice things about V: Numeric input is more convenient. Note how one input is an argument and the other is in the "input" field. This is an allowed input method, and it saves bytes because we want to run something n times, in this case the increment command or ctrl-a.

À           " Arg1 times:
 <C-a>      "   Increment the next number found on this line

Answer 19

SmileBASIC, 13 bytes

INPUT A,B?A+B

Answer 20

Python 3, 22 bytes

lambda a,b:a+b

Trivial answer for a kind of trivial question.

Answer 21

Wise, 9 bytes

[?~-!-~]|

Try it Online!

Takes 2 numbers in, A, and B.

             Implicit input
[ .... ]     While the top (B) != 0, repeat:
 ?           Move B to the bottom
  ~-         Add 1 to the top (A)
    !        Move B back to the top
     -~      Subtract 1 from the top (B)

        |    Gets rid of the 0 on top by ORing the top 2
             Implicit output

It boils down to "Add 1 to A, B times."

Less cool than the other Wise answer, but also shorter.

Answer 22

Braingolf, 1 byte

+

Try it online!

Answer 23

Triangular, 6 bytes

$.$%+<

Try it online!

Formats into this triangle:

  $
 . $
% + <

Without directionals or no-ops, the above turns into this: $$+%

Triangular uses postfix notation.

• $ - read input as integer
• + - add
• % - print as integer

Answer 24

,,,, 1 byte

+

^{Body must be at least 30 characters; you entered 14.}

Answer 25

Commentator, 4 bytes

////

Outputs by char code. If this isn't allowed, a 6 byte solution is

////*/

which outputs by numerical value.

Answer 26

tcl, 10

expr $a+$b

If you have namespace path tcl::mathop you can do it shorter for every arithmetic operation:

tcl, 7

+ $a $b

test case on: http://rextester.com/live/FLFPTH24568

Answer 27

Little Man Computer, 23 bytes (source)

INP
STA 6
INP
ADD 6
OUT

With HLT, 27 bytes (source)

INP
STA 6
INP
ADD 6
OUT
HLT

Online Emulator (Flash)

Answer 28

Bitwise, 89 bytes

IN 1 &1
IN 2 &1
LABEL &1
AND 1 2 *1
XOR 1 2 *2
SL *1 &1 1
MOV 2 *2 &1
JMP @1 *1
OUT *2 -1

Input and output are raw byte values as that is the only method of I/O for this language. Allowed per this consensus. Try it online!

Ungolfed, written properly, and commented:

IN 1 &1 -1       read a character into r1 if l1 truthy, discarding result
IN 2 &1 -1       read a character into r2
LABEL &1         create label 1 here
AND 1 2 *1       set fr1 to r1 & r2 where & represents bitwise AND
XOR 1 2 *2       set fr2 to r1 ^ r2 where ^ represents bitwise exclusive or
SL *1 &1 1       set r1 to fr1 << 1 where << represents bitwise left shift
MOV 2 *2 &1      move fr2 into r2 if &1 truthy
JMP @1 *1        jump to label 1 if fr1 truthy
OUT *2 &1 -1     print frame register 2 if l1 truthy, discarding result

Note that fr is short for frame register, r is short for register and l is short for literal.

Answer 29

Bash script, 12 bytes

expr $1 + $2

Save as add.sh, then run bash add.sh [argument] [argument].

Answer 30

NotQuiteThere, 8 bytes

0-- 0- 0

Try it online!

Given that one of the aims of this languages was to be unable to perform addition, I think I failed.

How it works

       # Implicit input; 10 and 20
0      # Push 0;   STACK = [10 20 0]
 -     # Subtract; STACK = [10 -20]
  -    # Subtract; STACK = [-30]
   0   # Push 0;   STACK = [-30 0]
    -  # Subtract; STACK = [30]
     0 # Push 0;   STACK = [30 0]
       # Output 30;